GENERAL I ARTICLE

The Jordan Curve Theorem 2. Conclusions

Ritabrata Munshi Introd uction

In the first part of the article (Resonance, Vol. 4, No.9 ) we proved the Jordan sepa.ration theorem which says that a simple closed curve in E2 separates it into at least two components. In this concluding part after some preliminary results we will prove the nonseparation theorem which states that an arc does not separate E2 and then put together Ritabrata Munshi is doing his M.Stat in the Indian all the pieces to conclude the proof of the Jordan Curve Statistical Institute, theorem. Calcutta. Apart from mathematics, he likes Prelhninary Lemmas painting and reading. We had defined when an arc is said to cross a circle. We Unlike most others he broaden the definition of crossing as follows: dislikes computers. Definition: Suppose f is a piece-wise circular simple closed curve and, is a piece-wise circular arc. Suppose we have the following situation: There exist PI, P2, P3, P4 E 'Y such that 'Y[P,P2] lies outside rand '[P:lP4] lies inside f while 'Y[P2,P3] is a subset of r (recall that JeT is true for r). Then we consider '[P2,P:l] as a crossing (see Figure 1).

It could happen that P2 = P3 in which case we are in the context of the earlier definition of crossing.

Lemma 2.1. Let 'Y, r be as above and p, q E 'Y such that P lies inside rand q lies outside r; then there is at least one crossing between l' and q.

Proof. Suppose 'Y : [0,1] -t E2, the map corresponding to 'Y. Let 'Y-1(1') rand ,-l(q) = s; without loss, assume Figure 1. = r < s. Let

0' = sup{x ::; s\'Y(x) E inside (f)};

then obviously 'Y ( a.) E rand ,(0: - f) lies inside of r for

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small f > O. Also, as q lies outside (f) there exists 5 such that I'(x) E r for 0. ~ X ~ 5 and 'Y(5 + E) E outside (f) for small f > O. Hence 'Y [0.,5] is a crossing.

So to 'move' from one component of (E2_f) to the other we have to cross f somewhere in between, and conversely, when­ ever we cross f we arrive at the other component. Hence Lemma 2.2. Let r be a piece-wise circular, simple, closed, curve, 'Y a piece-wise circular arc and a, b, the endpoints of 1'. If 'Y crosses r an odd number of times then a and bare in different components of (E2 - f).

Suppose I' is a polygonal arc and C is a circle, such that 'Y crosses C at least once and the end points, a and b, of I' lie 'outside' C. Let 'Ye and 'Yo be the exterior paths of 'Y w.r.t. C. (In the Part 1 of this article we had defined 'Ye and 1'0 when 'Y is a polygonal line. However, the same prescription works in the case when 'Y is a piece-wise circular arc.) Let 5e := ('Ye - 'Y) and 00 := (1'0 - 1'), so 5e and 0'0 are subsets of C. In fact 5e and 5'0 are finite disjoint union of open arcs of C. Let II, ,In be the sequence in which the (open) arcs are visited while moving from a to b; then 5e = Uk=lIk. Similarly, 50 = UbI Jl·

Lemma 2.3. Let p, q E oe and 0. be one of the arcs of C joining p and q. Then 0. contains an even number of crossings.

Proof. First note that as a, b lie outside C, the total number of crossings in C is even. If p, q E Ik for some k, then number of crossings is zero in one of the arcs of C and hence the result is true in this case.

N ow consider pElk, q E I k+ 1. If I k and I k+ 1 are separated by a point of touching (which is not a crossing) then the number of crossings is again zero in one direction.

In the other case, let 'Y(P/,ql] be the part of 'Y traveled while moving from Ik to Ik+1. Assume 'Y(p',q') n C = ¢ for the time being. If the direction at p is same as that at q then form the piece-wise circular simple closed curve 'Y[p',q'] Up'pqq' and call if ~. Now observe that exactly two situations can arise

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depending on whether C - p'pqq' lies outside or inside of (~):

(i) a, b are both inside (~) (Figure 2) or

(ii) a, b are both outside (~) (Figure 3).

Observe that in both cases the number of crossings in p'pqq' is even.

A similar argument proves the lemma if the directions at p Figure 2. and q are different.

Now if I(p'q) n C ¥= cp, then l'(p',q'] u p'pqq' will be a union of finite number of piece-wise circular simple closed curves. Then we can concentrate on each and every closed curve separately and prove the above claim.

So the statement is true for p E Ik and q E Ik+l. It is easy to see that this implies the general statement! (Why?)

Corollary 1. Suppose p E Dc, q E Dc and Q is an arc

of C joining p and q. Then Q ,contains an odd number of crossings.

Proof. It is enough to consider the case p E hand q E J1. And in this case a (the point where f hits the circle first) is the only crossing.

Corollary 2. oc U Oc = cp.

A Duality Result: Suppose TJ and ~ are two piece-wise circular simple closed curves and PI, P2 E TJ n ~ such that TJ 'crosses' ~ at TJ[Pl,P2] (= ~[Pl'P2])' then ~ also 'crosses' TJ at ~[Pl,P2]. (The notation TJ[pI,P2] and ~[Pl'P2] are ambiguous, because in each case there are two distinct arcs joining PI and P2. But here we take the arc common to TJ and ~ i.e. Figura 3. TJ[PIP2] = ~ (PIP2] . Also we have not defined crossing in this case; but the definition is obvious.) Proof. To prove duality we take sufficiently 'thin' strip neighbourhoods of TJ and ~ respectively and argue. Now we state and prove the main lemma of this section.

Lemma 2.4. Let TJ be an arc joining P E OC, q E Oc and

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Proof. Let 0: be an arc of C joining p and q. Let Rt, R2, ,R2s+1 be the crossings (of'Y and C) on 0:. Let Ca, Cb be the circles passing through a, b, respectively, concentric with C. We define a relation '-)0' on the set {Ri};!t1 by Ri --+ Rj if 'Y[R.i,Rj] does not interest CIl and Ca. Without loss, assume radius (Cb) ::; radius (Oa). It is easy to prove that '-)0' is an equivalence relation. It partitions the set {Rt, ,R2s+1} into equivalence classes. At least one of these equivalence classes has an odd number of points. Let {Ril' , Ri2J..+l} be one of them. Let E, F E Cb such that Ril' , Ri2>'+1 E 'Y(E,F) and 'Y(E,F) n Cb = ¢. Form ~ = 'Y[E,F] U EF (EF is anyone of the arcs of Cb joining E and

F.) Now ()i crosses it an odd number of times, hence p and q lie in different components. That 1] is an arc joining p and q implies ~ U 1] =1= ¢ but

1] n'Y = cp => 1] U E F =1= ¢ =} 1] U Cb =1= cp.

Also {p, q} C 1] U C, hence

diam(1]) ~ (radius Cb - radius C)

= min{ d(a, e), d(b, e)}. The N onseparation Theorem

Suppose 'Y : [0,1) '-)0 E2 is an arc and a, b are two points in (E2 - 'Y)' We will prove that there exists an arc joining a and b, not intersecting 'Y. As the proof is a bit technical, we will first present the basic plan:

Basic Plan: First we cover up 'Y with finite number of discs, this gives us a finite collection of circles. Suppose 1] in any piece-wise circular arc joining a and b, if it intersects 'Y then we can associate a number /-l(1], 'Y) with it,

N ow we choose a suitable circle C from the collection and deform 1} to 1]C such that either 1]C does not intersect 'Y or

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where 8 is a fixed positive number. Then we choose a subarc of 'fie joining a and b, and continue the process. As 8 is a fixed positive number ~ after finite number of deformations we will get a piece-wise circular arc not intersecting" but joining a and b.

N ow we give the proof in detail,

Proof. Let a := min{d(a, ,), d(b,,)} > O. As, is a home­ Figure 4. omorphism, for each p E , there exists p', p" 3 P E , (p' ,pi') and diam (,[p',pll]) < a/2. (Of course, at the end points we have to take p = p' and p =1= p".) As E2 is regular, for each p E , there exists an open disc B (p, Tp) not intersecting the closed set ,[o,p'] U ,[P1l,1] U {a, b}. (We denote ,(0) by 0 and , (1) by 1.) Again at the end points we have to make suit­ able modification, for example at ,(0), B(p, Tp) must avoid ,[pll ,1] U {a, b} only. Let Cp denote the boundary of B (p, T p). Now n = {B(p, T p/2) : p E ,} is an open cover of,. As , is compact there exists a finite sub cover

,t}, let e:= min{Tqd > O.

Now as , is uniformly continuous, there exists 8 > 0 such that

d(x,y) < 8 => d(,(x),,(y)) < e/2 V x,y E [0,1]

i. e. d (p, q) > e / 2 => d ( , -1 (p), , -1 (p)) > 8 V p, q, E , . ( *) Let 'fI be a piece-wise circular arc joining a and b, sup­ pose 'fI n, =1= ¢, then there exists io E {1, ,t} such that , (J.L ('fI, , )) E B (qio' , qiO/2)'

Let C denote the boundary of B (qio' Tqio ). Consider 'fie and 'fie; '[q:~,l] does not intersect them (as it does not intersect c and 'fI) and as diameter of the part of, from , ({t ('fI, , ) ) Figure 5. to q~~ is less than diam ,[q~o,q~~l < a/2, so that part does not intersect one of 'flO and 'fie (Lemma 2.4), say 'fie. So if 'fie intersects, then J.L (rye, ,) < J.L ('fI, , ). Moreover, (J.L ('fI, , ) ) lies inside B (qio' T q.iO/2 ~ and 'fie lies outside B (qio' T q'iO) hence

d(,(J.L('fI, ,)), rye) > Tqio/2 2: e

So using (*) we conclude J.L('fIO,,) < J.L('fI,') - 8.

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Proof of Jordan Curve Theorem

Let f be a simple closed curve in E2 and {rO:}O:EA be the components of (E2 - r). Now as r is topologically closed, each r 0: is open. Openness of r 0: assures us that A is a countable set.

Lemma 4.1 (i) Bd roC r for all a. E A;(ii) exactly one of r as has bounded complement.

Proof. (i) If possible choose p E (Bd r 0 - f). Now as r 0 is open p ¢ r o' Therefore there exists {3 -=1= a such that p E r{3. But this implies r 0: n r,a =I- ¢ (as rf3 is open), a contradiction. (ii) Take a circle C such that r lies inside (C). Then r 0: n J(C) 1= ¢ for all n. Now if r 0:0 n O(C) -=1= ¢ for some ao E A, then ro:o n C is non-empty. But C n f = ¢ implies C n Ed ro:o = ¢. Hence C C ro:o ~ O(C) c r 0:0'

Notation: 0 (r) = f 0:0 (the unique component with bounded

component) and J(r) = Uo:;iaor 0:' The non-separation theorem implies

Lemma 4.2. Bd ro: = r \f a.

Proof. Suppose Bd r 0: -=1= r for some a. As Ed r 0: is a closed set and B d r 0: c r there exists a subarc of r say ,/], such that Bd r 0: C 'TJ. Now, from the non-separation theorem 1] does not separate E2. Take pEr0: and q E r,B ({3 =I- a.), there exists an arc ~ (say), in (E2 - T}) joining p and q. Since p E ~ n r 0: and q E ~ n r~, this implies that ~ = (~n r 0:) u (~ U f'~) is a separation of ~. This contradicts the connectedness of ~.

Lemma 4.3. Suppose'Y and f are two simple closed curves in E2 and J (,) f- ¢. Then, intersects 0 (f) and J(f) implies that r intersects 0(,) and I('Y)'

2 Let '0: be a component of (E -,). We know that Bd ,0: = ,. Therefore, o(r)n, -=1= ¢ ~ o (f)n,o: -I ¢> ~ rn,o: -I ¢ (as '0: is connected).

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Figure 6 (left). Figure 7(right).

Therefore, r n 0(,) ::/= ¢. Arguing similarly, r n 1(,) i ¢. Now that we have all the necessary results, proof of the Jordan Curve Theorem can be quickly disposed off.

From the separation theorem it follows that f (r) =I- ¢ for any simple closed curve r We will prove that fer) is in fact a component of (E2-r). Hnot, then there exist two distinct components r a, r,a C fer). Take pEr0, we can find p',p" in r such that the line segments pp' and pp" excluding the points p' and pI! respectively, lie in r o. Similarly for a point q E 0 (r), we can find q', q" in r Let, be the simple closed curve formed by p' pp", q' qq", part of r between p' and q' and part of r between pI! and q", as shown (Figures 6 and 7). Then ,nO(r) f= ¢ and ,nf(r) f= rf> implies that rnO(,) f= ¢ and rnf(,) f= ¢. Hence O(,)nr0 and I(,)nr i= ¢ (Bd r,a = r). But r fJ n , = ¢ which contradicts the connectedness of r,a. Hence fer) is a component of (E2 - f). Thus a simple closed curve in the plane separates it into exactly two components, one bounded and the other un­ bounded.

Suggested Reading Address for Correspondence Ritabrata Munshi [1] J R Munkres, Topology -A first course, Prentice-Hall, Inc., 1983. Nivedita Sarani [2] John StiIlwell,Classical Topology and Combinatorial Group Theory,Springer Chandannagar Verlag, (GTM - 72), 1993. Hooghly 712136 [3] M A Armstrong, Basic Topology, Springer Verlag, (UTM), 1983. West Bengal. Indio.

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